MCQ
If $\sin^{-1}\text{x}-\cos^{-1}\text{x}=\frac{\pi}{6},$ then $x =$
  • A
    $\frac{1}{2}$
  • $\frac{\sqrt3}{2}$
  • C
    $-\frac{1}{2}$
  • D
    None of these

Answer

Correct option: B.
$\frac{\sqrt3}{2}$
We know that $\sin^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}$
$\therefore \sin^{-1}\text{x}-\cos^{-1}\text{x}=\frac{\pi}{6}$
$\Rightarrow\frac{\pi}{2}-\cos^{-1}\text{x}-\cos^{-1}\text{x}=\frac{\pi}{6}$
$\Rightarrow-2\cos^{-1}\text{x}=\frac{\pi}{6}-\frac{\pi}{2}$
$\Rightarrow-2\cos^{-1}\text{x}=-\frac{\pi}{3}$
$\Rightarrow\cos^{-1}\text{x}=\frac{\pi}{6}$
$\Rightarrow\text{x}=\cos\frac{\pi}{6}$
$\Rightarrow\text{x}=\frac{\sqrt3}{2}$

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